A "well-known" fact is that a ${C}_q[U(1)]$-coaction on a vector space $V$ induces a $Z$-grading. I don't see why this is so. Clearly, a ${C}_q[U(1)]$-grading will induce a $Z$-grading. But why does a ${C}_q[U(1)]$-coaction induce a ${C}_q[U(1)]$-grading? In other words, Why should every $v \in V$ be equal to a sum $\sum_i v_i$, for which $\Delta (v_i) = v_i \otimes k^n$?

Moreover, what happens when we "move up" to a ${C}_q[U(2)]$-coaction, or a ${C}_q[U(N)]$-coactions? Does this induce a ${C}_q[U(2)]$-grading, and a ${C}_q[U(N)]$-grading?

What is $C_q (U(N))$? Coactions by functions on a group are trivially grading but I am not sure that I understood the question.
–
Bugs BunnyDec 3 '10 at 8:59

$C_q[U(N)]$ is the quantised (noncommutative) coordinate algebra of $U(N)$. The definition is similar to that of the better known $C_q[SU(N)]$ but with one extra generator det$^{-1}$ and the relation det.det$^{-1} = $det$^{-1}$.det$=1$ instead of det$=1$.
–
Dyke Acland Dec 3 '10 at 15:48

1 Answer
1

This follow from the axioms of a coaction, and the Hopf algebra structure of $C_q[U(1)]$. Firstly, for $v \in V$, we will always have $\Delta_R(v) = \sum_i v_i \otimes k^{n_i}$. Applying the equality id$ \otimes \varepsilon = $id to both sides gives $v = \sum_i v_i$. We now use the axiom $(\Delta_R \otimes $id$) \circ \Delta_R = ($\id$ \otimes \Delta) \circ \Delta_R$. Applying both sides to $v$ we get $\sum_{ij} v_{ij} \otimes k^{n_j} \otimes k^{n_i} = \sum v_i \otimes k^{n_i} \otimes k^{n_i}$. Now consider the set of elements which are the sums of the terms with the same $k^{n_i}$ as the third tensor factor. This set is linearily independent. Thus, $\sum_j v_{ij} \otimes k^{n_j} = f_i \otimes k^{n_i}$. Since $\Delta_R (f_i) = \sum_j f_{ij} \otimes k^{n_j}$, we have the required result.